122 
PROFESSOR HUGH L. CALLENDAR ON 
In order to calculate the value of this correction in terms of the heat-loss and the 
flow, it is necessary to consider the differential equation of the distribution of tem¬ 
perature in the flow-tube. We assume, as a first approximation, that the temperature 
of the surface of the flow-tube, on which the loss of heat depends, is at each point 
nearly the same as that of the liquid flowing through it. This is very nearly true in 
the case of mercury, and to a sufficient approximation in the case of water under 
suitable conditions. 
The rate of evolution of heat by a current C amperes in a length dx of the tube is 
C 2 rdx watts, where r is the resistance in ohms per centim. The rate of loss of heat 
is fdp dx, where p is the perimeter of the flow-tube in centims., and f the emissivity 
in watts per sq. centim. per degree of temperature excess d. The rate of gain of heat 
by the liquid is JsQ dd, where J is the number of joules in one calorie ; s the specific 
11 eat of the liquid in calories per gramme degree C ; Q the flow of liquid in grammes 
per second; and dd the rise of temperature in a length dx. Since it is necessary to 
take account of the change of resistance with temperature, we must substitute for r 
the value r Q (1 + ad), where a is the temperature-coefficient of the increase of 
resistance, and r 0 the value of r when 0=0. We thus obtain the linear differential 
equation 
J.sQ dd/dx -f- (fp — C 2 cir 0 ) d = CV 0 .( 1 ). 
Writing for brevity, A = ( fp — C 2 «r 0 ) / JhsQ, and B = CV 0 /JsQ, and observing 
that d — 0 when x — 0 , since the liquid flows into the tube at the jacket-temperature, 
the solution of this equation is 
d = (l - <r Ax ) B/A.( 2 ). 
Since A is very small compared with B in the actual experiment, we may obtain a 
sufficient approximation for our purpose by expanding the exponential and neglecting 
the terms beyond A 2 , which gives 
d x — B.r (1 — Ax/2) 
If the whole length of the fine flow-tube is l, the temperature at the end of the 
flow-tube will be approximately 
0i = Bl (1 - M/2) .(4), 
and the mean temperature d m from 0 to /, will be 
dm = B/ (1 - AZ/3)/2.(5). 
It will be observed that the value of A is zero, and that the gradient is constant 
throughout the tube for a particular value of the current, C 2 = fp/cir 0 , which may be 
called the critical value of the current. In this case the radiation-loss along the 
